For representing floating point numbers, we use float, double and long double.
What’s the difference ?
double has 2x more precision then float.
float is a 32 bit IEEE 754 single precision Floating Point Number1 bit for the sign, (8 bits for the exponent, and 23* for the value), i.e. float has 7 decimal digits of precision.
double is a 64 bit IEEE 754 double precision Floating Point Number (1 bit for the sign, 11 bits for the exponent, and 52* bits for the value), i.e. double has 15 decimal digits of precision.
Let’s take a example(example taken from here) :
For a quadratic equation x2 – 4.0000000 x + 3.9999999 = 0, the exact roots to 10 significant digits are, r1 = 2.000316228 and r2 = 1.999683772
// C program to demonstrate // double and float precision values #include <stdio.h> #include <math.h> // utility function which calculate roots of // quadratic equation using double values void double_solve(double a, double b, double c){ double d = b*b - 4.0*a*c; double sd = sqrt(d); double r1 = (-b + sd) / (2.0*a); double r2 = (-b - sd) / (2.0*a); printf("%.5f %.5f
", r1, r2); } // utility function which calculate roots of // quadratic equation using float values void float_solve(float a, float b, float c){ float d = b*b - 4.0f*a*c; float sd = sqrtf(d); float r1 = (-b + sd) / (2.0f*a); float r2 = (-b - sd) / (2.0f*a); printf("%.5f %.5f
", r1, r2); } // driver program int main(){ float fa = 1.0f; float fb = -4.0000000f; float fc = 3.9999999f; double da = 1.0; double db = -4.0000000; double dc = 3.9999999; printf("roots of equation x2 - 4.0000000 x + 3.9999999 = 0 are :
"); printf("for float values:
"); float_solve(fa, fb, fc); printf("for double values:
"); double_solve(da, db, dc); return 0; } |
Output:
roots of equation x2 - 4.0000000 x + 3.9999999 = 0 are : for float values: 2.00000 2.00000 for double values: 2.00032 1.99968
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